Question

a.. Differentiate the function 𝑦 = (3𝑥^2+2) ^2√6𝑥+2 / 𝑥 ^3+1

b. Let P and Q be points on the curve 𝑦 = 𝑥 2 − 2𝑥 while x = 2 and x = 2 + h respectively. Express the gradient of pq in terms of h and hence find the gradient of the curve 𝑦 = 𝑥 2 − 2𝑥 at x= 2

c. Find the gradient of the curve 𝑦 = 1/ 𝑥−1 at the point (2,1)

A straight line 𝑦 = −𝑥 + 4 cuts the parabola with equation 𝑦 = 16 − 𝑥 2 at the points A and B.

a) Find the coordinates of A and B

b) Calculate the distance between the points A and B

c) Find the equation of the tangents at A and B, and hence determine where the tangents meet.

d) The line µ is perpendicular to the line A at the point A and B meet. Give its equation

Answer 1

1.

a.

$$y'=((3x^2+2)^2\sqrt{6}x+\dfrac{2}{x^3+1})'$$

$$=12\sqrt {6}x^2(3x^2+2)+\sqrt{6}(3x^2+2)^2-\dfrac{6x^2}{(x^3+1)^2}$$

b.

$$grad=\dfrac{(2+h)^2-2(2+h)-2^2+2(2)}{2+h-2}$$

$$=\dfrac{4+4h+h^2-4-2h-4+4}{h}=2+h$$

$$\lim\limits_{h\to 0}\dfrac{(2+h)^2-2(2+h)-2^2+2(2)}{2+h-2}$$

$$=\lim\limits_{h\to 0}(2+h)=2+0=2$$

Gradient of the curve $y=x^2-2x$ at $x=2$ is 2.

c.

$$y'=(\dfrac{1}{x-1})'=-\dfrac{1}{(x-1)^2}$$

$$grad|_{(2, 1)}=-\dfrac{1}{(2-1)^2}=-1$$

2.

a) Find the coordinates of A and B

$$-x+4=16-x^2$$

$$x^2-x-12=0$$

$$(x+3)(x-4)=0$$

$x_1=-3, y_1=-(-3)+4=7$

$x_2=4, y_2=-4+4=0$

$A(-3, 7), B(4, 0)$

b)

$$d_{AB}=\sqrt{(4-(-3))^2+(0-7)^2}=2\sqrt{7}$$

c)

$$(16-x^2)'=-2x$$

$A(-3, 7)$

$$y'(-3)=-2(-3)=6$$

$$y-7=6(x-(-3))$$

The equation of the tangent line

$$y=6x+25$$

$B(4, 0)$

$$y'(4)=-2(4)=-8$$

$$y-0=-8(x-4)$$

The equation of the tangent line

$$y=-8x+32$$

$$6x+25=-8x+32$$

$$14x=7$$

$$x=\dfrac{1}{2}$$

$y=-8(\dfrac{1}{2})+32=28$

The tangents meet at the point $(\dfrac{1}{2}, 28)$

d)

The line µ is perpendicular to the line A is perpendicular to the line $y=-x+4$

$$slope=1$$

$$y=x+b$$

The line µ passes through the point $(\dfrac{1}{2}, 28)$

$$28=\dfrac{1}{2}+b=>b=\dfrac{55}{2}$$

The equation of the line µ is $y=x+\dfrac{55}{2}.$